On 2012-04-11 23:32 +0300, Rémi Denis-Courmont wrote: > From the perspective of the compiler, this is a feature not a bug. In > C and C++, loading or storing a value in an enumerated variable > whereby the value is not a member of the enumeration is undefined. I'm afraid that this is not the case in C, although it may be in C++ (enums are very different in C++ than they are in C). In C, enum types are required to be compatible with some integer type capable of storing the values of all the enum members (see C11§6.7.2.2#4). Compatibility is a very strong condition, and implies that the two types are interchangable without affecting the meaning of the program (see C11§6.2.7). Integer types have a number of specific requirements, one thing that's relevant here is that they do not have "holes" in their representable values: there is a minimum and maximum representable value, and all integers between them are representable (C11§6.2.6.2#1). Thus, while the choice of integer type used may depend on the values of the corresponding enum constants, storing any value (regardless of whether or not its a member of the enumeration) is subject to the same rules as the implementation-defined compatbile integer type. This is always well-defined for values within the range of the type. (C11§6.3.1.3#1 and C11§6.3.1.4#1). > In other words, the compiler can assume that this does not happen, and > optimize it away. No, a conforming C compiler cannot assume such assignments do not happen, for the reasons outlined above. Cheers, -- Nick Bowler, Elliptic Technologies (http://www.elliptictech.com/) -- To unsubscribe from this list: send the line "unsubscribe linux-media" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
